I have a shuffled deck of $d$ cards containing $g$ “good” cards. I'm going to draw a hand of $h$ cards (without replacement) and I want it to contain exactly $w$ good cards. (Change the variables if there's a standard notation.) What is the probability of drawing a hand with the cards I want?
I feel like I knew a general solution to this in high school, but I can't find it now. (I can find some specific solutions, like this one.)
What I've got is this:
$$\frac{\binom{g}{w} \cdot \binom{d-g}{h-w}}{\binom{d}{h}} = \frac{\frac{g!}{w! \cdot (g-w)!} \cdot \frac{(d-g)!}{(h-w)! \cdot ((d-g)-(h-w))!}}{\frac{d!}{h! \cdot (d-h)!}} = \frac{g! \cdot (d-g)! \cdot h! \cdot (d-h)!}{w! \cdot (g-w)! \cdot (h-w)! \cdot (d-g-h+w)! \cdot d!}$$
which is way more complicated than I expected, but is giving the answers I expect.
E.g.: Draw 5 cards from a standard deck and want all 4 aces:
$$\frac{\binom{4}{4} \cdot \binom{52-4}{5-4}}{\binom{52}{5}} = 0.000018$$
Am I missing something? Is there an easier way? Is there at least a simpler simplification?
In case that's the end of that question, bonus question!
If I deal a hand like this to $p$ players, what is the probability that one (or at least one) of them gets a qualifying hand?
$w$ is the count of favored items in a sample of size $h$ selected without bias from a population of size $d$ containing $g$ favoured items. The disribution for this is known as the Hypergeometric Distribution. It does not get any easier than this.
Well, the probability for selecting $w$ from $g$ cards when drawing $h$ from $d$ cards equals the probability for putting $w$ of the good cards into $h$ special places when sorting all $g$ good cards among $d$ positions. But that other way of viewing the task is not any simpler to evaluate.
$$\dfrac{\dbinom gw\dbinom{d-g}{h-w}}{\dbinom dh}=\dfrac{\dbinom{h}{w}\dbinom{d-h}{g-w}}{\dbinom dg}$$
The task might be laborious by hand, but scientific calculators (devices or apps) usually provide a binomial function.